Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $a = \dfrac{t^2 - 7t}{t^2 - 6t - 40} \times \dfrac{8t + 32}{t - 7} $
Explanation: First factor the quadratic. $a = \dfrac{t^2 - 7t}{(t + 4)(t - 10)} \times \dfrac{8t + 32}{t - 7} $ Then factor out any other terms. $a = \dfrac{t(t - 7)}{(t + 4)(t - 10)} \times \dfrac{8(t + 4)}{t - 7} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ t(t - 7) \times 8(t + 4) } { (t + 4)(t - 10) \times (t - 7) } $ $a = \dfrac{ 8t(t - 7)(t + 4)}{ (t + 4)(t - 10)(t - 7)} $ Notice that $(t - 7)$ and $(t + 4)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ 8t(t - 7)\cancel{(t + 4)}}{ \cancel{(t + 4)}(t - 10)(t - 7)} $ We are dividing by $t + 4$ , so $t + 4 \neq 0$ Therefore, $t \neq -4$ $a = \dfrac{ 8t\cancel{(t - 7)}\cancel{(t + 4)}}{ \cancel{(t + 4)}(t - 10)\cancel{(t - 7)}} $ We are dividing by $t - 7$ , so $t - 7 \neq 0$ Therefore, $t \neq 7$ $a = \dfrac{8t}{t - 10} ; \space t \neq -4 ; \space t \neq 7 $